Problem 34
Question
Compute the directional derivative of \(f(x, y)\) at the point \(P\) in the direction of the point \(Q\). $$ f(x, y)=e^{x-y}, P=(2,2), Q=(1,-1) $$
Step-by-Step Solution
Verified Answer
The directional derivative is \( \frac{\sqrt{10}}{5} \).
1Step 1: Find the Gradient Vector
To find the directional derivative, first calculate the gradient vector \( abla f(x, y) \) of the function \( f(x, y) = e^{x-y} \). The partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = e^{x-y} \), and the partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = -e^{x-y} \). Thus, the gradient vector is \( abla f(x, y) = (e^{x-y}, -e^{x-y}) \).
2Step 2: Evaluate the Gradient at Point P
Substitute the point \( P = (2,2) \) into the gradient vector to evaluate it at this point. This gives \( abla f(2,2) = (e^{2-2}, -e^{2-2}) = (e^0, -e^0) = (1, -1) \).
3Step 3: Find the Direction Vector from P to Q
Determine the direction vector by subtracting the coordinates of \( P = (2,2) \) from \( Q = (1,-1) \). The direction vector is \( \mathbf{d} = (1 - 2, -1 - 2) = (-1, -3) \).
4Step 4: Normalize the Direction Vector
Calculate the magnitude of the direction vector \( \mathbf{d} = (-1, -3) \), which is \( \| \mathbf{d} \| = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \). The unit vector in the direction of \( \mathbf{d} \) is \( \hat{\mathbf{d}} = \left(-\frac{1}{\sqrt{10}}, -\frac{3}{\sqrt{10}}\right) \).
5Step 5: Compute the Directional Derivative
The directional derivative \( D_{\mathbf{u}}f(P) \) is the dot product of \( abla f(2,2) = (1, -1) \) and the unit direction vector \( \hat{\mathbf{d}} = \left(-\frac{1}{\sqrt{10}}, -\frac{3}{\sqrt{10}}\right) \). Perform the dot product: \[ D_{\mathbf{u}}f(P) = 1 \cdot \left(-\frac{1}{\sqrt{10}}\right) + (-1) \cdot \left(-\frac{3}{\sqrt{10}}\right) = -\frac{1}{\sqrt{10}} + \frac{3}{\sqrt{10}} = \frac{2}{\sqrt{10}}. \]To simplify, we can rationalize the denominator: \[ \frac{2}{\sqrt{10}} \times \frac{\sqrt{10}}{\sqrt{10}} = \frac{2\sqrt{10}}{10} = \frac{\sqrt{10}}{5}. \]
Key Concepts
Gradient VectorPartial DerivativeUnit Vector
Gradient Vector
The gradient vector is a key concept in multivariable calculus. It represents the direction of the steepest ascent of a function. For a function with two variables, like in this exercise with \( f(x, y) = e^{x-y} \), the gradient vector is composed of two partial derivatives. It can be represented as \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). The partial derivative with respect to \( x \) (\( \frac{\partial f}{\partial x} \)) gives us how the function changes as \( x \) changes while keeping \( y \) fixed, and similarly for \( y \).
In this exercise, we found the gradient as \( abla f(x, y) = (e^{x-y}, -e^{x-y}) \). At the point \( P = (2, 2) \), it evaluates to \( abla f(2, 2) = (1, -1) \). This means that moving in the direction of (1, -1) from \( P \) will decrease the function value the fastest.
Understanding the gradient vector is crucial as it serves as a foundational tool in finding directional derivatives as well as in optimization problems.
In this exercise, we found the gradient as \( abla f(x, y) = (e^{x-y}, -e^{x-y}) \). At the point \( P = (2, 2) \), it evaluates to \( abla f(2, 2) = (1, -1) \). This means that moving in the direction of (1, -1) from \( P \) will decrease the function value the fastest.
Understanding the gradient vector is crucial as it serves as a foundational tool in finding directional derivatives as well as in optimization problems.
Partial Derivative
Partial derivatives are used to measure how a multivariable function changes as one variable varies, with all other variables held constant. If you have a function \( f(x, y) \), the partial derivative with respect to \( x \), symbolized as \( \frac{\partial f}{\partial x} \), tells us the rate at which \( f \) changes as \( x \) changes, while \( y \) is constant.
In this problem, the partial derivative with respect to \( x \) is \( e^{x-y} \), and similarly, with respect to \( y \) is \( -e^{x-y} \). Each of these partial derivatives contributes to forming the gradient vector, which plays a crucial role in finding the directional derivative. Getting a good grasp on partial derivatives helps in understanding complex surfaces in 3-dimensional space, shaping the path to understanding phenomena in physics and engineering.
Breaking down a function into partial derivatives is a powerful technique to predict how small changes in input variables influence the output, an important part of multivariable calculus.
In this problem, the partial derivative with respect to \( x \) is \( e^{x-y} \), and similarly, with respect to \( y \) is \( -e^{x-y} \). Each of these partial derivatives contributes to forming the gradient vector, which plays a crucial role in finding the directional derivative. Getting a good grasp on partial derivatives helps in understanding complex surfaces in 3-dimensional space, shaping the path to understanding phenomena in physics and engineering.
Breaking down a function into partial derivatives is a powerful technique to predict how small changes in input variables influence the output, an important part of multivariable calculus.
Unit Vector
A unit vector is a vector of length one that indicates direction. When calculating the directional derivative, it’s vital to express any direction vector as a unit vector to ensure the directional derivative accurately reflects the rate of change per unit distance in that direction.
Here, our direction vector from the point \( P = (2, 2) \) to \( Q = (1, -1) \) was \( (-1, -3) \). To convert this into a unit vector, first calculate its magnitude using the formula \( \| \mathbf{d} \| = \sqrt{(-1)^2 + (-3)^2} = \sqrt{10} \). Then, divide the direction vector by its magnitude to normalize it: \( \hat{\mathbf{d}} = \left(-\frac{1}{\sqrt{10}}, -\frac{3}{\sqrt{10}}\right) \).
Understanding unit vectors is essential because it standardizes the direction, making comparisons and further calculations more manageable, especially in physics and engineering, where direction matters as much as magnitude.
Here, our direction vector from the point \( P = (2, 2) \) to \( Q = (1, -1) \) was \( (-1, -3) \). To convert this into a unit vector, first calculate its magnitude using the formula \( \| \mathbf{d} \| = \sqrt{(-1)^2 + (-3)^2} = \sqrt{10} \). Then, divide the direction vector by its magnitude to normalize it: \( \hat{\mathbf{d}} = \left(-\frac{1}{\sqrt{10}}, -\frac{3}{\sqrt{10}}\right) \).
Understanding unit vectors is essential because it standardizes the direction, making comparisons and further calculations more manageable, especially in physics and engineering, where direction matters as much as magnitude.
Other exercises in this chapter
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